Let $C>0$ and $(X_n)$ be a sequence of positive random variables. Assume that $$ |X_n - C| = o_p(r_n^{-1}) \iff r_n|X_n-C|=o_p(1) $$ for some fixed sequence $(r_n)$ with $r_n \to \infty$.
What can we say about the rate of convergence of the log-transform: $$ |\log(X_n)- \log(C)| = o_p(?). $$ I guess it depends on $C$ and $(r_n)$ but I can't seem to derive anything useful.
It has the same rate.
I hope that the following hint can help.
Recall that: $r_n|X_n-C|=o_p(1) \iff \left( \forall \varepsilon>0, \mathbb P(|X_n-C|>\frac{\epsilon}{r_n}) \rightarrow 0 \text{ when } n\rightarrow \infty \right)$.
For the case $X_n>C+\frac{\epsilon}{r_n}$, we apply the Taylor expansion and get $$r_n[\log(X_n)-\log(C)]> \frac{\varepsilon}{C} + O(\varepsilon^2).$$
Similar for the case $X_n<C-\frac{\epsilon}{r_n}$.
Edit. As pointed out by @John, above implication is not correct, what we should prove is the reversion. However, the same idea can be applied, i.e, we should use Taylor expansion of $e^x$ instead of $\log x$.
In the following, I propose a more detailed solution without using Taylor expansion (to avoid big-O notation).
Claim: For all $\varepsilon>0$, exists $\delta>0$ such that $$r_n|\log(X_n)-\log(C)|>\varepsilon \Longrightarrow r_n|X_n-C|> \delta$$ for $n$ large enough.
Proof of Claim: Let $Y_n=\log X_n$ and $B=\log C$. So we want to show that, for all $\varepsilon>0$, exists $\delta>0$ such that $$r_n|Y_n-B|>\varepsilon \Longrightarrow r_n|e^{Y_n}-e^B|> \delta$$ for large $n$.
Now assume that $r_n|Y_n-B|>\varepsilon$, we consider two cases.
First, recall a simple inequality that $e^{x}\geq 1+x$ for all $x\in \mathbb R$. If $Y_n > B +\frac{\varepsilon}{r_n}$, then $e^{Y_n} > e^B(1 +\frac{\varepsilon}{r_n})$ and thus $r_n(e^{Y_n} - e^B)>e^B \varepsilon$.
Second, it is easy to prove that $e^{-x} \leq 1-x+\frac{x^2}{2}$ for $x\geq 0$. If $Y_n < B -\frac{\varepsilon}{r_n}$, then $r_n(e^{Y_n} - e^B)<-e^B \varepsilon+e^B\frac{\varepsilon^2}{2r_n}$.
Third, note that for any $\varepsilon>0$, there exists a sufficiently large integer $n$ such that $\frac{\varepsilon}{2r_n}<1$. Choose $M$ such that $\frac{\varepsilon}{2r_n}<M<1$ and $\delta:=e^B(1-M)\varepsilon$.
Then, we have $r_n|e^{Y_n} - e^B|>e^B \varepsilon-e^B\frac{\varepsilon^2}{2r_n}> e^B(1-M)\varepsilon=\delta$. Done.